Why WeakReference still return the Object's Value?

Question

I wanna know how to use WeakReference in developing android application.So as i searched a lot in the internet i typed this code :

   {
        String kk = "Test";
        WeakReference<String> kkRef = new WeakReference<String>(kk);
        System.out.println(kk);
        System.out.println(kkRef.get());
        kk = null;
        System.gc();
        System.out.println(kk);
        System.out.println(kkRef.get());
   }

which suppose to eligible the kk object to GC, But in the console output it seems that kkRef.get() still return the "Test" value which supposed to return null as i read about weakReference

So Why it didn't return null?

CommonsWare · Accepted Answer · 2016-05-01 20:13:15Z

2

First, System.gc() does not necessarily perform a full garbage collection run, and there is no guarantee that it will do so synchronously. Hence, it is not guaranteed to detect all weakly-held objects.

Second, the string is still held onto strongly, as it is a string literal in your code. That cannot be garbage collected ever, and so a WeakReference to it would still return the literal.

answered May 1, 2016 at 20:13

CommonsWare

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4 Comments

Mohamed Fadel Buffon Over a year ago

But i typed kk = null which remove the strong reference

CommonsWare Over a year ago

@MohamedFadelBuffon: You removed that strong reference. The string literal still exists in your code. In Java, assignment (String kk = "Test") does not make a copy of the object. The string literal cannot be garbage collected. Try using an instance of an object that you create yourself using the new keyword.

Mohamed Fadel Buffon Over a year ago

Thank u very much it works for me :) . But could u please expalin to me what you meant by literal .. Thanks in advance

CommonsWare Over a year ago

@MohamedFadelBuffon: A literal in Java is any primitive value (5, 3.14159f, true) or String ("Test") that you enter directly into your Java code.

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